Question

When $22.4L$ of $H_2(g)$ is mixed with $11.2L$ of $C{l}_2(g)$, each at $STP$, the moles of $HCl(g)$ formed is equal to

• A. 1 mole of HCI (g)
• B. 2 moles of HCI (g )
• C. 0.5 mole of HCI (g)
• D. 1.5 moles of HCI (g)

Answer 1

Answer: (A)

The given problem is related to the concept of stoichiometry of chemical equations. Thus, we have to convert the given volumes into their moles and then, identify the limiting reagent [possessing minimum number of moles and gets completely used up in the reaction]. The limiting reagent gives the moles of product formed in the reaction.

$\because 22.4L$ volume at STP is occupied by
$C{l}_2=1$ mole
$\therefore 11.2L$ volume will be occupied by
$C{l}_2=\frac{1\times 11.2}{22.4}mol=0.5mol$
$22.4L$ volume at STP is occupied by $H_2=1mol$

Since, $C{l}_2$ possesses minimum number of moles, thus it is the limiting reagent.
As per equation,
$1$ mole of $C{l}_2=2$ moles of $HCl$
$\therefore 0.5$ mole of $C{l}_2=2\times 0.5$ mole of $HCl$
$=1.0$ mole of $HCl$
Hence, $1.0$ mole of $HCl(g)$ is produced by $0.5$ mole of $C{l}_2$ [or $11.2L$ ].